Let $n$ an integer sufficiently large.

I'm looking for a "big" set $S$ of prime number $p$ satisfying $10.n \geq p > \sqrt{18n} $ and $p| C_{9.n}^{k-n}C_{8n+k}^{8.n}$ for all integer $k$ satifying $ n \leq k \leq 10n$ and $k \geq p $

that means I'm looking a heavy condition $(C)$ over the $p$ prime $10n \geq p > \sqrt{18n} $ giving me for all $p$ prime satisfying $(C)$ we have for all integers $k$ such that $\max(n,p) \leq k \leq 10n$ one has

$$v_p(C_{9n}^{k-n}C_{8n+k}^{8n})=\Big[\frac{8n+k}{p}\Big]-\Big[\frac{k}{p}\Big]-\Big[\frac{8n}{p}\Big]+\Big[\frac{9n}{p}\Big]-\Big[\frac{k-n}{p}\Big]-\Big[\frac{10n-k}{p}\Big] \geq 1$$

where $[x]$ is the integer part of a real number $x$.

I think $(C) $ is of type $p \geq \sqrt{18n}, \frac{a}{b}<\frac{n}{p}-[\frac{n}{p}]< \frac{c}{d} $for some integer $a,b,c,d $ that I coudn't find, which gives us the existence of an integer $h$ such that $p \in \Big]\displaystyle \frac{dn}{c+dh}, \displaystyle \frac{nb}{a+rh}\Big[$.

I have proved that for all $p \in(8n,9n),$ one has $v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$. There are many others prime number satisfying $v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$, maybe in $\big(\frac{8n}{2},\frac{9n}{2}\big), \big(\frac{8n}{3},\frac{9n}{3}\big)...$.

I don't know what the $a,b,c,d$ are. Numerically, I have found many $p$ satisfying $v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$ and $10n \geq p > \sqrt{18n}$.

Any help is appreciated.

Let $n$ an integer sufficiently large.

I'm looking for a "big" set $S$ of prime number $p$ satisfying $10.n \geq p > \sqrt{18n} $ and $p| C_{9.n}^{k-n}C_{8n+k}^{8.n}$ for all integer $k$ satifying $ n \leq k \leq 10n$ and $k \geq p $

that means I'm looking a heavy condition $(C)$ over the $p$ prime $10n \geq p > \sqrt{18n} $ giving me for all $p$ prime satisfying $(C)$ we have for all integers $k$ such that $\max(n,p) \leq k \leq 10n$ one has

$$v_p(C_{9n}^{k-n}C_{8n+k}^{8n})=\Big[\frac{8n+k}{p}\Big]-\Big[\frac{k}{p}\Big]-\Big[\frac{8n}{p}\Big]+\Big[\frac{9n}{p}\Big]-\Big[\frac{k-n}{p}\Big]-\Big[\frac{10n-k}{p}\Big] \geq 1$$

where $[x]$ is the integer part of a real number $x$.

I think $(C) $ over the prime number satisfying $10n \geq p \geq \sqrt{18n}$ is:

$$p \in \displaystyle {\cup}_{h=1}^8 ]\displaystyle \frac{8n}{h},\displaystyle \frac{9n}{h}[ $$

I have proved that for all $p $ satifying $ 10.n \geq p > \sqrt{18n} $ and in $]8n,9n[,$ one has $v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$. There are many others prime number satisfying $v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$, maybe in $\big(\frac{8n}{2},\frac{9n}{2}\big), \big(\frac{8n}{3},\frac{9n}{3}\big)...$.

Numerically, I have found many $p$ $10n \geq p > \sqrt{18n}$ satisfying $ \forall k $ such that $10n \geq k \geq p$ we have $ v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$ and $10n \geq p > \sqrt{18n}$.

Any help is appreciated.

let $n$ an integer sufficiently large.

i'm looking for a "big" set $S$ of prime number $p$ satisfying $10.n \geq p > \sqrt{18n} $ and $p| C_{9.n}^{k-n} .C_{8n+k}^{8.n}$ for all integer $k$ satifying $ n \leq k \leq 10.n$ and $k \geq p $

that means I'm looking a havy condition $(C)$ over the $ p $ prime $10.n \geq p > \sqrt{18n} $ giving me for all $p$ prime satisfying $(C)$ we have

$\forall k$ integer such that $max(n,p) \leq k \leq 10.n$

$$ (I): v_p(C_{9.n}^{k-n} .C_{8n+k}^{8.n})=[\frac{8.n+k}{p}]-[\frac{k}{p}]-[\frac{8.n}{p}]+[\frac{9.n}{p}]-[\frac{k-n}{p}]-[\frac{10.n-k}{p}] \geq 1$$

where $[x]$ is the integer part of a real number $x$

( i think $(C) $ is of type $p \geq \sqrt{18.n}, \frac{a}{b}<\frac{n}{p}-[\frac{n}{p}]< \frac{c}{d} $for some integer $a,b,c,d $ that i coudn't find, wich is give us to the existence of an integer $h$ such that $p \in ]\displaystyle \frac{d.n}{c+d.h}, \displaystyle \frac{n.b}{a+r.h}[ $ )\

i prooved that $ \forall p \in]8.n,9.n[,$ i have $(I)$ there are many others prime number satisfying $(I)$ may be in $]\frac{8n}{2},\frac{9n}{2}[, ]\frac{8n}{3},\frac{9n}{3}[...$ i don't know what are the $a,b,c,d$ i numerically i have many $p$ satisfying $(I)$ and $10.n \geq p > \sqrt{18n} $... Any help please

Let $n$ an integer sufficiently large.

I'm looking for a "big" set $S$ of prime number $p$ satisfying $10.n \geq p > \sqrt{18n} $ and $p| C_{9.n}^{k-n}C_{8n+k}^{8.n}$ for all integer $k$ satifying $ n \leq k \leq 10n$ and $k \geq p $

that means I'm looking a heavy condition $(C)$ over the $p$ prime $10n \geq p > \sqrt{18n} $ giving me for all $p$ prime satisfying $(C)$ we have for all integers $k$ such that $\max(n,p) \leq k \leq 10n$ one has

$$v_p(C_{9n}^{k-n}C_{8n+k}^{8n})=\Big[\frac{8n+k}{p}\Big]-\Big[\frac{k}{p}\Big]-\Big[\frac{8n}{p}\Big]+\Big[\frac{9n}{p}\Big]-\Big[\frac{k-n}{p}\Big]-\Big[\frac{10n-k}{p}\Big] \geq 1$$

where $[x]$ is the integer part of a real number $x$.

I think $(C) $ is of type $p \geq \sqrt{18n}, \frac{a}{b}<\frac{n}{p}-[\frac{n}{p}]< \frac{c}{d} $for some integer $a,b,c,d $ that I coudn't find, which gives us the existence of an integer $h$ such that $p \in \Big]\displaystyle \frac{dn}{c+dh}, \displaystyle \frac{nb}{a+rh}\Big[$.

I have proved that for all $p \in(8n,9n),$ one has $v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$. There are many others prime number satisfying $v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$, maybe in $\big(\frac{8n}{2},\frac{9n}{2}\big), \big(\frac{8n}{3},\frac{9n}{3}\big)...$.

I don't know what the $a,b,c,d$ are. Numerically, I have found many $p$ satisfying $v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$ and $10n \geq p > \sqrt{18n}$. 

Any help is appreciated.